Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 5x}{x + 6} = \dfrac{50}{x + 6}$
Solution: Multiply both sides by $x + 6$ $ \dfrac{x^2 + 5x}{x + 6} (x + 6) = \dfrac{50}{x + 6} (x + 6)$ $ x^2 + 5x = 50$ Subtract $50$ from both sides: $ x^2 + 5x - (50) = 50 - (50)$ $ x^2 + 5x - 50 = 0$ Factor the expression: $ (x - 5)(x + 10) = 0$ Therefore $x = 5$ or $x = -10$ The original expression is defined at $x = 5$ and $x = -10$, so there are no extraneous solutions.